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 Algebra Pre-Algebra and Basic Algebra Math Forum

 June 5th, 2014, 02:48 AM #1 Senior Member   Joined: Nov 2013 Posts: 434 Thanks: 8 log log15/log6=a log18/log12=b find log25/log24
 June 5th, 2014, 04:54 AM #2 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,156 Thanks: 731 Math Focus: Physics, mathematical modelling, numerical and computational solutions $\displaystyle a = \frac{\log 15}{\log6} = \frac{\log 3 + \log 5}{\log 3 + \log 2}$ $\displaystyle b = \frac{\log 18}{\log 12} = \frac{\log 2 + 2\log 3}{\log 3 + 2\log 2}$ Rearrange a to make $\displaystyle \log 5$ the subject: $\displaystyle \log 5 = a(\log 3 + \log 2) - \log 3$ Rearrange b to make $\displaystyle \log 3$ the subject: $\displaystyle \log 3 = \frac{2b-1}{2-b} \cdot \log 2$ $\displaystyle \frac{\log 25}{\log 24}= \frac{2\log 5}{3\log 2 + \log 3}$ $\displaystyle = \frac{2a(\log 3 + \log 2) - 2\log 3}{3\log 2 + \log 3}$ $\displaystyle = \frac{2a\left(\frac{2b-1}{2-b}\log 2 + \log 2\right) - 2\left(\frac{2b-1}{2-b}\log 2\right)}{3\log 2 + \left(\frac{2b-1}{2-b}\log 2\right)}$ $\displaystyle = \frac{2a\left(\frac{2b-1}{2-b}\log 2\right) + 2a\log 2 - 2\left(\frac{2b-1}{2-b}\log 2\right)}{3\log 2 + \left(\frac{2b-1}{2-b}\log 2\right)}$ $\displaystyle = \frac{2a\left(\frac{2b-1}{2-b}\right) + 2a - 2\left(\frac{2b-1}{2-b}\right)}{3 + \left(\frac{2b-1}{2-b}\right)}$ $\displaystyle = \frac{2a(2b-1) + 2a(2-b) - 2(2b-1)}{3(2-b) + 2b-1}$ $\displaystyle = \frac{4ab -2a+4a-2ab-4b+2}{6-3b+2b-1}$ $\displaystyle = \frac{2ab+2a-4b+2}{5-b}$ $\displaystyle \frac{2(ab+a-2b+1)}{5-b}$

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